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CHAPTER 11. SUPERPOSITION OF LIGHT WAVE
When two waves meet at a particular point in space, the resultant disturbance is
simply the algebraic sum of the constituent disturbance. This is the principle of
superposition. In mathematics, particularly, if the wavefunctions 1 and 2 are each
separate solutions of the wave equation, it follows that (1+2) is also a solution. This is
the linear superposition of waves.
11.1. Addition of Waves of the Same Frequency
Suppose we have two wavefunctions,
 1  E1 sin kx  t   1 
 2  E2 sin kx  t   2 
(11.1)
(11.2)
where 1 and 2 are the initial phases at x=0 and t=0. For simplicity, let,
1  kx   1
 2  kx   2
and
(11.3)
Hence we have,
 1  E1 sin 1  t 
 2  E2 sin  2  t  (11.4)
and
The resultant wave can be expressed as,
  1  2  E sin   t   E sin kx  t   
(11.5)
where,
E 2  E12  E22  2E1 E2 cos 2  1 
(11.6)
E sin 1  E2 sin  2
tan   1
(11.7a)
E1 cos 1  E2 cos  2
E sin  1  E2 sin  2
tan   1
(11.7b)
E1 cos  1  E2 cos  2
Since the intensity of a wave is proportional to the amplitude squared, from
Eq.(11.6) we can obtain,
I  I1  I 2  2 I1 I 2 cos 2  1 
(11.8)
The term 2 I1 I 2 cos 2  1  is known as the interference term.
When 2-1=2m, where m is any integer or zero, the two waves are said to be in
phase. The interference is called the total constructive interference and yields the
maximum intensity,
I max  I1  I 2  2 I1 I 2
(11.9)
When 2-1=(2m+1), the two waves are said to be antiphase or 180 out-ofphase. The interference is called the total destructive interference and yields the
minimum intensity,
I min  I1  I 2  2 I1 I 2
(11.10)
Waves for which the initial phase difference 2-1 is constant are said to be coherent.
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Coherent waves have a constant spatial relationship between them. If the initial phase
difference is varies randomly with time, then the waves are said to be incoherent and the
intensity of the resultant waves is the sum of the intensity of each wave.
For waves at distances x1 and x2 from their sources, the phase difference of the two
waves are given by,
  kx2  t   2   kx1  t   1   k x2  x1    2   1  (11.11)
For waves in a medium of refractive index n, m=/n, the phase difference will be,
2
x2  x1    2   1   2 nx2  x1    2   1 

(11.12)
m

The quantity n(x2-x1) is called the Optical Path Difference and is abbreviated to OPD.
For a material of refractive index n with a thickness t, the number of wavelengths in
the thickness t is,
t
nt

(11.13)
 n 
The term nt is the optical thickness or Optical Path Length (OPL). If the wave travels
several different materials, the OPL will be the sum of the individual OPLs. The number
of wavelengths in this optical path length is (OPL)/.
The superposition of waves can be done using the phasor diagram (Fig.11.1).
Suppose we have two wave functions,
 1  E01 sin t  1  and  2  E02 sin t   2 
(11.14)
The resultant wave function is,
   1   2  E0 sin t   
(11.15)
where,
E0 
E01 cos 1  E02 cos  2 2  E01 sin 1  E02 sin  2 2
tan  
E01 sin 1  E02 sin  2
E01 cos 1  E02 cos  2
(11.16)
(11.17)
Fig.11.1. Phasor diagram for the superposition
of two harmonic waves. Each wave can be
represented by a vector with a magnitude
equal to the amplitude of the wave. The
vector forms between the positive x-axis an
angle equal to the phase angle .
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The graphical procedure can be extended to accommodate any number of
component waves of the same frequency. To generalize,
N
   E0i sin t   i   E0 sin t   
i 1
E0 
X 2 Y2
tan   Y X
and
N
where, X   E 0i cos  i
i 1
(11.18)
(11.19)
N
and
Y   E0i sin  i
i 1
(11.20)
Example: Find the resultant of adding the sine waves
 1  20 sin t
 3  10 sin t   12
 2  10sin t   4
 4  15sin t  2 3
Example: Find, using algebraic addition, the amplitude and phase resulting from the
addition of the two superposed waves  1  E1 sin kx  t  1  and
 2  E2 sin kx  t   2 , where 1=0, 2=/2, E1=8, E2=6, and x=0.
Example: Two waves  1  E1 sin kx  t  and  2  E2 sin kx  t    are coplanar and
overlap. Calculate the resultant’s amplitude if E1=3 and E2=2.
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Example: Show that the optical path length, or more simply the optical path, is equivalent
to the length of the path in vacuum which a beam of light of wavelength  would
traverse in the same time.
Let’s consider two harmonic waves of the same frequency propagating in opposite
directions,
 I  E0 I sin kx  t   I 
and  R  E0 R sin kx  t   R 
Assuming E0I=E0R, I=R=0, the resultant wave is then
   1  2  2E0 I sin kx cos t
(11.21)
This is the equation for standing or stationary wave, as opposed to traveling wave. Its
profile does not move through space. At any fixed point, the amplitude is a constant equal
to 2 E0 I sin kx and varies harmonically as cos t. At certain points, namely, x=0, /2, ,
3/2,…, the disturbance will be zero at all times. These are known as nodes or nodal
points. Halfway between each adjacent node, that is, at x=/4, 3/4, 5/4,…, the
amplitude has a maximum value and these points are known as the antinodes.
Fig.11.2. A standing wave
at different times.
11.2. Addition of Waves of Different Frequency
Let’s consider the addition of two waves differing in frequency and wave number,
 1  E1 cosk1 x  1t 
 2  E1 cosk2 x  2t 
and
but with equal amplitudes and zero initial phases.
The superposition of these two waves is
   1  2
   2   k1  k 2
  2 
k  k
 2 E1 cos  1 2 x  1
t  cos 
x 1
t
2
2
 2
  2

 2 E1 cosk p x   p t cosk g x   g t 
(11.22)
The above equation represents a product of two cosine waves. The first possesses a
frequency p and propagation number kp that are, respectively, the averages of the
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frequencies and propagation numbers of the component waves. The second cosine factor
represents a wave with a modulation frequency g and modulation propagation number kg
that are much smaller by comparison.
With p >> g , plots of the composite wave appear like those of Fig.11.3. The
overall effect is that the low-frequency wave serves as an envelope modulating the highfrequency wave. Eq.(11.22) represents a traveling wave of frequency p and propagation
number kp with the amplitude given by,
E  2 E1 cosk g x   g t 
(11.23)
Therefore, when the composite wave is traveling, the shape of the wave is kept on
modulated. The rate at which the modulation envelope (as shown in Fig.11.3) advances is
known as the group velocity.
 g 1   2 
vg 


(11.24)
kg
k1  k 2
k
Fig.11.3. (Left) Separate plot of the wave function where p >> g. (Right) the composite wave.
Advanced Study:
The composite wave shown in Fig.11.3 exhibits the phenomenon of beats. The figure shows that
the beat frequency is twice the frequency of the modulating envelop, or
 b  2 g  1   2
If more than two sinusoidal waves of different frequencies are added to give a group
of waves. The changing profile of these superposed waves is called a wave packet, wave
group or wave train. If the difference in frequency is small, the group velocity can be
expressed as,
d
vg 
(11.25)
dk
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In ordinary media  is dependent on . The particular function =(k) is called a
dispersion relation.
Advanced Study:
The modulation or signal propagates at a speed vg that may be greater than, equal to, or less than
v, the phase velocity. Since =kv, Eq.(11.25) yields,
dv
vg  v  k
dk
Given that n(k) is known, v=c/n, and it is useful to reformulate the group velocity as,
c kc dn
vg   2
n n dk
 k dn 
v g  v 1 
or

 n dk 
For normal dispersion where dn/dk>0, the group velocity is less than the phase velocity. For
anomalous dispersion where dn/dk<0, the group velocity is larger the than phase velocity. In vacuum
where there is no dispersion, the group velocity is equal to the phase velocity.
Example: Suppose a group of electromagnetic waves with frequencies from 41014 Hz to
51014 Hz propagates in vacuum. Find the group velocity.
In a wave packet, the frequency bandwidth is defined as    1  2 , where 1
and 2 are the two extreme frequencies of the group. From Fourier analysis, the frequency
bandwidth is approximately equal to the reciprocal of the coherent time,
1
t 
(11.26)

For waves propagating in vacuum with speed c (3108m/s), the length of the wave packet
is known as the coherent length,
x  ct
(11.27)
Example: (a) How many vacuum wavelengths of =500 nm will span space of 1 m in a
vacuum?
(b) How many wavelengths span the gap when the same gap has a 10 cm thick slab
of glass (ng=1.5) inserted in it?
(c) Determine the optical path difference between the two cases.
(d) Verify that OPD/ is the difference between the answers to (a) and (b).
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Example: In the figure, two waves 1 and 2 both have vacuum wavelengths of 500 nm.
The waves arise from the same source and are in phase initially. Both waves travel
an actual distance of 1 m but
2 passes through a glass tank
with 1 cm thick walls and a
20 cm gap between the walls.
The tank is filled with water
(nw=1.33) and the glass has
refractive index ng=1.5. Find
the OPD and the phase
difference when the waves
have traveled the 1 m distance.
Example: Show that the standing wave s(x,t) is periodic with time. That is, show that
s(x,t)= s(x,t+).
Exercises:
11.1 Add the two coplanar waves 1=Esin(kx-t) and 2=3Ecos(kx-t-/2).
11.2 Add the two coplanar waves 1=3sin(kx-t) and 2=4sin(kx-t+/4). What is the
phase difference between the waves 1 and 2?
11.3 A light wave, =660 nm, passes into and out of a glass tank which has walls 5 cm
thick, an internal length of 50 cm, and ng=1.6. If the glass is replaced by a plastic
material of refractive index 1.5, what wall thickness will be required to keep the
optical path length the same assuming the internal length remains the same? How
many waves occupy the wall-interior-wall length?
11.4 Which two of the following waves will interfere to produce (a) total constructive
interference and (b) total destructive interference?
 1  4 sin kx  t 
 2  4 coskx  t   2
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 3  4 coskx  t   2
 4  4 sin kx  t   4
What is the amplitude of the resultant wave in (a)?
11.5 Two coherent coplanar waves have amplitudes of 2 and 3 units and the same
frequency, but a phase difference of /3. Find the intensity of the resultant wave
relative to that of the individual waves. If the waves are incoherent, what then is the
resultant intensity relative to the individual waves?
11.6 Suppose a group of electromagnetic waves with frequencies from 41014 Hz to
51014 Hz propagates in a medium for which the refractive indices are n1=1.49 and
n2=1.51, respectively. (a) Calculate the group velocity in the medium. (b) Calculate
the phase (or wave) velocities of the two extreme components. (c) Using the
extreme frequencies, calculate the coherence time and the coherence length.
11.7 A ‘particle’ of light, a photon, is emitted by an atom in about 10 -8 s. How long is
the wave packet?
References:
1.
A.H.Tunnacliffe and J.G.Hirst, Optics, The Association of British Dispensing
Opticians, 1996.
2.
F.L.Pedrotti and L.S.Pedrotti, Introduction to Optics, Prentice Hall, New Jersey,
1993.
3.
E.Hecht, Optics, Addison Wesley, San Francisco, 2002.
4.
D.Halliday, R.Resnick and J.Walker, Fundamentals of Physics, 5th edition, John
Willey & Sons, Inc., Singapore, 1997.
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